The owner of a fish market has an assistant who has determined that the weights of catfish are normally​ distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 64 fish yields a mean of 3.4​ pounds, what is probability of obtaining a sample mean this large or​ larger? Round to four decimal places.

Answer:

39.20000

Step-by-step explanation:

Sweater costs 56.00

To calculate the discounted price of the sweater

56 * (30/100) = 16.8

We used (30/100) to find out the 30 percent of the sweater which will be on discount/reduced.

56-16.8=39.2 is the discounted price

Then we subtract the amount of discount from the original price of sweater to find out the discounted price.

Answer: switch the x- and y- coordinates

And the 2nd part is c, e , f

The point on the graph -1(x) =9X to determine a point on the graph of f(x) =logox is (-1, -9).

All the points (and only those points) which lie on the graph of the function satisfy its equation.

Thus, if a point lies on the graph of a function, then it must also satisfy the function.

We are given that;

-1(x) =9X

Now,

To use a point on the graph of one function to determine a point on the graph of another function, you need to find the corresponding x-value on the first graph. Then, you can use that x-value to find the corresponding y-value on the second graph.

In this case, you have a point on the graph of the function f(x) = 9x. To find the corresponding point on the graph of the function g(x) = log(x), you need to find the x-value that corresponds to -1 on the graph of f(x) = 9x.

To do this, you can set f(x) = 9x equal to -1 and solve for x:

9x = -1 x = -1/9

The point (-1, -9) on the graph of f(x) = 9x corresponds to the point (-1/9, log(-1/9)) on the graph of g(x) = log(x). However, note that the logarithm function is not defined for negative values of x, so the point (-1/9, log(-1/9)) is not a valid point on the graph of g(x) = log(x).

Therefore, by the graph of function the point will be (-1, -9).

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Answer:

The proportion of jugs which receive more than 65 ounces of detergent is 0.0062 or 0.62%

Step-by-step explanation:

Mean amount of detergent = u = 64

Standard deviation = [tex]\sigma[/tex] = 0.4

We need to find the proportion of jugs with over 65 ounces of detergent. Since the population is Normally Distributed and we have the value of population standard deviation, we will use the concept of z-score to solve this problem.

First we will convert 65 to its equivalent z-score, then using the z-table we will desired proportion. The formula to calculate the z-score is:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

x = 65 converted to z score will be:

[tex]z=\frac{65-64}{0.4}=2.5[/tex]

Therefore, the probability of detergent being more than 65 ounces is equivalent to probability of z-score being over 2.5

i.e.

P(X > 65) = P(z > 2.5)

From the z-table and using the property of symmetry:

P(z > 2.5) = 1 - P(z < 2.5)

= 1 - 0.9938

= 0.0062

Therefore,

P(X > 65) = P(z > 2.5) = 0.0062

So, the proportion of jugs which receive more than 65 ounces of detergent is 0.0062 or 0.62%

To find the proportion of jugs that receive more than 65 ounces of detergent when the machine is running on target, we can use the properties of the normal distribution. The proportion is about 0.62%.

To find the proportion of jugs that receive more than 65 ounces of detergent when the machine is running on target, we can use the properties of the normal distribution. Since the mean amount is 64 ounces and the standard deviation is 0.4 ounces, we can calculate the Z-score for 65 ounces using the formula Z = (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation.

Plugging in the values, we get Z = (65 - 64) / 0.4 = 2.5. Now, we can look up the cumulative probability for a Z-score of 2.5, which represents the proportion of jugs that receive less than or equal to 65 ounces. Subtracting this proportion from 1 gives us the proportion of jugs that receive more than 65 ounces of detergent when the machine is running on target.

Using a Z-table or a calculator, we find that the cumulative probability for a Z-score of 2.5 is approximately 0.9938. Subtracting this from 1, we find that the proportion of jugs that receive more than 65 ounces of detergent when the machine is running on target is about 0.0062 or about 0.62%.

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Answer:

As Null hypothesis is not satisfied so there is no evident that nutrition label doesn't provide accurate measure of calories.

Answer:

Yes

Step-by-step explanation:

Complete question is:

The nutrition label on a bag of potato chips says that a one ounce(28g) serving of potato chips has 130 calories and contains 10 grams of fats with 3 grams of saturated fats. A random sample of 35 bags yielded a confidence interval for the number of calories per bag of 128.2 to 139.8 calories. Is there evidence that the nutrition label does not provide an accurate measure of calories in the bags of potato chips?

The calories stated in nutriton label (130) is close to the lower bound of confidence interval range which is 128.2. So this can be an evidence that nutrition lable may not provide an accurate measure of calories in bags.  

Answer:

The answer is 115/1957 and in decimal form is 0.0558634

Step-by-step explanation:

Answer:

y = -2x+3

Step-by-step explanation:

Answer:

We have to produce the equation y = mx +b

We start by solving for b

b = y - mx

b = 3 - -2*0

b = 3

Let's put b and the slope into this equation

y = mx +b

y = -2*x +3

Step-by-step explanation:

Answer:

-A person can be 95% confident that the mean drive through service time  lies between 177.6 seconds and 181.0 seconds.

Step-by-step explanation:

- Confidence level is the degree of certainty we have on a particular statistic.

-The 95% confidence interval means that we are 95% confident that the mean drive through service time is lies between 177.6 seconds and 181.0 minutes.

Answer:

A quadratic function can have two irrational roots.

A quadratic equation can have no real number solutions.

Step-by-step explanation:

A quadratic function can have two irrational roots.

A quadratic equation can have no real number solutions.

A quadratic function is a polynomial function with one or more in which highest exponent of variable is two.

According to the question,

A quadratic function can have two irrational roots.

example: [tex]x^{2} -2x -2 =0[/tex]

This equation have two irrational roots  1+√3 and 1 - √3.

A  quadratic function can have no real number solutions

Hence, A quadratic function can have two irrational roots.

A quadratic equation can have no real number solutions.

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The function f(x) = 4x2 + x4 + 3 is even because the substitution of (-x) for x results in the original function. It's not odd because replacing (-x) for x doesn't give the negative of the original function. Hence, as an even function, its graph is symmetric with respect to the y-axis.

The function f(x) = 4x2 + x4 + 3 can be tested for symmetry. If a function is even, its graph is symmetric with respect to the y-axis. If a function is odd, its graph is symmetric with respect to the origin.

To test if a function is even, we substitute (-x) for x in the function and simplify. If the result is the original function, then the function is even. For the given function, f(-x) = 4(-x)2 + (-x)4 + 3 = 4x2 + x4 + 3. So, the function is even.

To test if a function is odd, we also substitute (-x) for x in the function and simplify. If the result is the negative of the original function, then the function is odd. In our case, f(-x) is not the negative of f(x), so the function is not odd.

Therefore, the function is even and its graph is symmetric with respect to the y-axis.

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Answer:

3.78% probability that exactly five cars will arrive over a five-minute interval during rush hour

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

20 cars per 10 minutes

So for 5 minutes, [tex]\mu = 10[/tex]

What is the probability that exactly five cars will arrive over a five-minute interval during rush hour?

This is P(X = 5).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378[/tex]

3.78% probability that exactly five cars will arrive over a five-minute interval during rush hour

The probability that exactly five cars will arrive over a five-minute interval during rush hour is approximately 0.0378 or 3.78%.

Firstly, the arrival rate is given as 20 cars per 10 minutes. Thus, the arrival rate per minute, λ, is 20 cars / 10 minutes = 2 cars per minute. To find the arrival rate for a 5-minute interval, we multiply this rate by 5 minutes: λ = 2 cars/minute * 5 minutes = 10 cars.

The Poisson probability formula is:
P(X = k) = (λ^k * e^(-λ)) / k!
where λ is the average number of cars in the interval, k is the number of cars, and e is the base of the natural logarithm (approximately equal to 2.71828).

In this problem, we need to find the probability of exactly 5 cars arriving in a 5-minute interval. Thus, λ = 10 and k = 5:

P(X = 5) = (10^5 * e^(-10)) / 5!
P(X = 5) = (100000 * e^(-10)) / 120
P(X = 5) ≈ (100000 / 148.4132) / 120
P(X = 5) ≈ 0.0378

Therefore, the probability that exactly five cars will arrive over a five-minute interval during rush hour is approximately 0.0378 or 3.78%.

Answer:

  year 2139

Step-by-step explanation:

The population will double when the factor e^(.005t) is 2.

  e^(.005t) = 2

  .005t = ln(2)

  t = ln(2)/0.005 = 138.6

The population will be double its size at t=0 when t=138.6. That is the population will be about 5.2 million in the year 2139.

The population will double by the year 2139 from its value of 2.6 million in year 2000.

Population function :

[tex]P(t) = 2.6 {e}^{0.005t} [/tex]

Population size at t = 0

[tex]P(0) = 2.6 {e}^{0.005(0)} = 2.6(1) = 2.6[/tex]

Population at t = 2.6 million.

For the population to double ;

2.6 × 2 = 5.2 million :

[tex]5.2 = 2.6 {e}^{0.005t} [/tex]

We solve for t

[tex] \frac{5.2}{2.6} = {e}^{0.005t} [/tex]

[tex]2 = {e}^{0.005t} [/tex]

Take the In of both sides

[tex] ln(2) = 0.005t[/tex]

[tex]t \: = ln(2) \div 0.005 = 138.629[/tex]

The population will double after 139 years

Therefore, the population will double by the 2139 (Year 2000 + 139 years) = year 2139.

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Answer:

56,000

Step-by-step explanation:

In your first year of work, you would make $56,000.

The pay rate is $28 per hour and a work week has 40 hours for the HVAC technician.

In a week, you will make:

= Number of hours in week x Amount per hour

= 40 x 28

= $1,120

In a year, assuming there are 50 weeks, the HVAC technician would make:

= Amount per week x Number of weeks in year

= 1,120 x 50

= $56,000

In conclusion, you would make $56,000 a year as an HVAC technician.

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Answer:

a) Discrete, because the number of point scored during basket ball is countable.

For instance, the amount of point scored in a basketball could be 75, 103, 63 etc. The numbers are countable

b) Continuous, because the amounts of rainfall is a random variable that is uncountable.

For instance, the amount of rainfall in City Upper B during April could be 0.10 inches of rain per hour, 0.30 inches of rain per hour. This numbers are not countable, they are rather approximated or rounded off.

Step-by-step explanation:

A random variable is considered discrete if its possible values are countable while a random variable is considered to be continuous if it's possible values are not countable.

Answer:

[tex]10.08-1.64\frac{0.1}{\sqrt{6}}=10.013[/tex]    

[tex]10.08+1.64\frac{0.1}{\sqrt{6}}=10.147[/tex]    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma =0.1[/tex] represent the population standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=10.08[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Now we have everything in order to replace into formula (1):

[tex]10.08-1.64\frac{0.1}{\sqrt{6}}=10.013[/tex]    

[tex]10.08+1.64\frac{0.1}{\sqrt{6}}=10.147[/tex]    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

Answer:

1271.7 cm3

Step-by-step explanation:

Answer:

1,271.7cm‬

Step-by-step explanation:

V=π*r^2 * (h/3)

V= 3.14  * 9^2  * (15/3)

V= 3.14  * 81 * 5  =1271.7

Answer:

Ans: n = [1.645/0.02]^2*0.13*0.87 = 766 when rounded up

Step-by-step explanation:

Answer:

I hope this helps

Answer:

We have,

max 4x+10y

3x+4y<480

4x+2y<360

The feasible corner points are ( 48,84),(0,120),(0,0),(90,0)

Now, our problem is maximum type so put above feasible points in equation max 4x + 10y one by one and select one point at which our value from this equation is maximum,

(48,84) 4*48+10*84 1032

(0,120) 4*0+10*120 1200

(0,0) 4*0+10*0 0

(90,0) 4*90+10*0 360

Here we get maximum value at (0,120) which is 1200.

Correct option is (B)1200

Answer:

Answer is 1200.

Refer below.

Step-by-step explanation:

The maximum possible value for the objective function is 1200.

Answer:

a

Step-by-step explanation:

a yard is 3 feet.24/3 equals 8 .then 80!

Answer:

Step-by-step explanation:

Hello!

According to a study regarding the average age of female and male kids to complete toilet training is:

Females:

Average age 32.5 months

The standard deviation of 6.7 months

n= 126

Males:

Average age 35.0 months

The standard deviation of 10.1 months

n= 141

The parameter of study is the difference between the age of females are successfully toilet trained and the average age that males are successfully toilet trained. μf - μm (f= female and m= male)

a.

Assuming that both variables have a normal distribution to choose whether you'll use an unpooled or pooled-t to calculate the confidence interval you have to conduct an F-test for variance homogeneity.

If the variances are equal, then you can usee the pooled-t, but if the variances are different, you have to uses Wlche's approach:

H₀: δ²f = δ²m

H₁: δ²f ≠ δ²m

Since both items b. and c. ask for a 95% CI I'll use the complementary significance level for this test:

α: 0.05

[tex]F= \frac{S^2_f}{S^2_m} * \frac{xSigma^2_f}{Sigma^2_m} ~~~F_{(n_f-1); (n_m-1)}[/tex]

[tex]F= \frac{(6.7^2)}{(10.1)^2} *1= 0.44[/tex]

Critical values:

[tex]F_{125;140;0.025}= 0.71\\F_{125;140;0.975}= 1.41[/tex]

The calculated F value is less than the lower critical value, 0.77, so the decision is to reject the null hypothesis. In other words, there is no significant evidence to conclude the population variances of the age kids are toilet trained to be equal. You should use Welch's approach to construct the Confidence Intervals.

[tex]Df_w= \frac{(\frac{S^2_f}{n_f} + \frac{S^2_m}{n_m} )^2}{\frac{(\frac{S^2_f}{n_f} )^2}{n_f-1} +\frac{(\frac{S^2_m}{n_m} )^2}{n_m-1} }[/tex]

[tex]Df_w= \frac{(\frac{6.7^2}{126} + \frac{10.1^2_m}{141} )^2}{\frac{(\frac{126^2}{126} )^2}{126-1} +\frac{(\frac{10.1^2}{141} )^2}{141-1} } = 254.32[/tex]

b.

The given interval is:

[0.3627; 4.6073]

Using Welch's approach, the formula for the CI is:

(X[bar]f- X[bar]m) ± [tex]t_{Df_w;1-\alpha /2}[/tex] * [tex]\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }[/tex]

or

(X[bar]m- X[bar]f) ± [tex]t_{Df_w;1-\alpha /2}[/tex] * [tex]\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }[/tex]

As you can see either way you calculate the interval, it is centered in the difference between the two sample means, so you can clear the value of that difference by:

(Upper bond - Lower bond)/2= (4.6073-0.3627)/2= 2.1223

The average age for females is 32.5 months and for males, it is 35 months.

Since the difference between the sample means is positive, we can say that the boys were considered "group 1" and the girls were considered "group 2"

You have a95% confidence that the parameter of interest is included in the given confidence interval.

c.

None of the statements is correct, the interval gives you information about the difference between the average age the kids are toilet trained, that is between the expected ages for the entire population of male and female babies.

This represents a guideline but is not necessarily true to all individuals of the population since some male babies can be toilet trained before that is expected as some female babies can be toiled trained after the average value.

I hope it helps!

Answer:

[tex]\left[\begin{array}{ccc}\dfrac{1}{2} &\dfrac{1}{2}\\\\\dfrac{1}{4}&\dfrac{3}{4}\end{array}\right][/tex]

Step-by-step explanation:

Let 1 = Sorey State, and 2 = C&T

Half the owners of Sorey State Boogie Boards became disenchanted with the product and switched to C&T Super Professional Boards the next surf season.

Three quarters of the C&T Boogie Board users remained loyal to C&T, while the rest switched to Sorey State.

The Markov Transition Matrix is presented below:

[tex]\left\begin{array}{ccc}\\\\\\$Sorey State&1\\\\C\&T&2\end{array}\right\left[\begin{array}{ccc}$Sorey State&C\&T\\1&2\\------&------\\\dfrac{1}{2} &\dfrac{1}{2}\\\\\dfrac{1}{4}&\dfrac{3}{4}\end{array}\right][/tex]

The above is presented for clarity sake. The transition matrix is:

[tex]\left[\begin{array}{ccc}\dfrac{1}{2} &\dfrac{1}{2}\\\\\dfrac{1}{4}&\dfrac{3}{4}\end{array}\right][/tex]

The Markov transition matrix, based on the given question, would look as follows: The top row represents the switch from Sorey State to C&T (0.5) and C&T to Sorey State (0.25). The bottom row represents the loyal customers who stick with their Sorey State (0.5) and C&T (0.75) boards. This matrix represents the probability of customers transitioning between these two brands in one surf season.

To properly answer this question, we need to convert these figures into a Markov transition matrix. In a Markov transition model, each consumer either stays with the brand they have (represented by the numbers on the diagonals) or switches to the other brand (represented by the numbers not on the diagonals).

Given the problem, set it up as follows:

As a matrix, this looks as follows:

[Sorey State, C&T Boards]

[0.5, 0.25]

[ 0.5, 0.75]

This is your completed Markov transition matrix, which represents the probability of customers transitioning between these two brands, from one surf season to the next.

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Answer:

63720

Step-by-step explanation:

yes

Answer:

The base is a square of side 9.19 cm and the height is 7.66 cm

[tex]C_m=126.58\ cents[/tex]

Step-by-step explanation:

Optimization

We'll use simple techniques to find the optimum values that minimize the cost function given in the problem. Since the restriction is an equality, the derivative will come handy to find the critical points and then we'll prove they are a minimum.

First, we consider the shape of the rectangular bin has a square base and no top. Let x be the side of the base, thus the Area of the base is

[tex]A_b=x^2[/tex]

Let y be the height of the box, thus each one of the four lateral sides of the box is a rectangle with sides x and y and the total lateral area is

[tex]A_s=4xy[/tex]

The cost of the material used to manufacture the box is 0.5 cents per square centimeter of the base and 0.3 cents per square centimeter of the sides, thus the total cost to produce one box is

[tex]C(x,y)=0.5x^2+0.3\cdot 4xy[/tex]

[tex]C(x,y)=0.5x^2+1.2xy[/tex]

Note the cost is a two-variable function. We need to have it expressed as a single variable function. To achieve that, we use the volume provided as [tex]646 cm^3[/tex]. The volume of the box is the base times the height

[tex]V=x^2y[/tex]

Using the value of the volume we have

[tex]x^2y=646[/tex]

Solving for y

[tex]\displaystyle y=\frac{646}{x^2}[/tex]

Replacing into the cost function, it only depends on one variable

[tex]\displaystyle C(x)=0.5x^2+1.2x\cdot \frac{646}{x^2}[/tex]

Operating

[tex]\displaystyle C(x)=0.5x^2+ \frac{775.2}{x}[/tex]

Taking the first derivative

[tex]\displaystyle C'(x)=x-\frac{775.2}{x^2}[/tex]

Equating to 0

[tex]\displaystyle x-\frac{775.2}{x^2}=0[/tex]

Solving

[tex]\displaystyle x=\sqrt[3]{775.2}[/tex]

[tex]x=9.19\ cm[/tex]

Now find the height

[tex]\displaystyle y=\frac{646}{9.19^2}[/tex]

[tex]y=7.66\ cm[/tex]

Find the second derivative

[tex]\displaystyle C''(x)=1+\frac{1550.4}{x^3}[/tex]

Since this value is positive, for all x positive, the function has a minimum at the critical point.

Thus, the minimum cost is

[tex]\displaystyle C_m=0.5\cdot 9.19^2+ \frac{775.2}{9.19}[/tex]

[tex]\boxed{C_m=126.58\ cents}[/tex]

Answer:

126.58 cents or $1.27

Step-by-step explanation:

the math from above is correct they just want the answers in dollars

Answer:Y=0 Y=-3

Step-by-step explanation:

Answer:

First graph y=0

Second graph y=-3

Step-by-step explanation:

Edge 2022

Answer:

try y=3/4x+2

Step-by-step explanation:

2 because it is the y intercept

3/4 because it is the slope

Answer:

The slope for this is 3/4. y=3/4x+2

Answer:

1) Increase the sample size

2) Decrease the confidence level

Step-by-step explanation:

The 95% confidence interval built for a sample size of 1100 adult Americans on how much they worked in previous week is:

42.7 to 44.5

We have to provide 2 recommendations on how to decrease the margin of Error. Margin of error is calculated as:

[tex]M.E=z_{\frac{\alpha}{2} } \times \frac{\sigma}{\sqrt{n}}[/tex]

Here,

[tex]z_{\frac{\alpha}{2} }[/tex] is the critical z-value which depends on the confidence level. Higher the confidence level, higher will be the value of critical z and vice versa.

[tex]\sigma[/tex] is the population standard deviation, which will be a constant term and n is the sample size. Since n is in the denominator, increasing the value of n will decrease the value of Margin of Error.

Therefore, the 2 recommendations to decrease the Margin of error for the given case are:

The two recommendations should be that the sample size should be increased and the confidence interval should be reduced.

Since a​ 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5.

We know that margin of error = z value × population / √n

So for reducing the margin of error of the interval,  sample size should be increased and the confidence interval should be reduced.

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Answer:

you are to subtract the equation

Answer:

(7,-8)

Step-by-step explanation:

Answer:??

Step-by-step explanation:

Answer: dont have an answer

Step-by-step explanation:

sorry